- #1
mikelepore
- 551
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Can someone please tell me why the formula for B at the center of a coil and the formula for B inside a solenoid give widely different answers in this problem?
From Serway and Vuille, College Physics: "A solenoid with 500 turns, 0.10 m long, carrying a current of 4.0 A and with a radius of 10^-2 m will have what strength magnetic field at its center?" -- Answer provided by the publisher: 250 X 10^-4 T
My comparison of the two formulas ...
Method 1
Using the formula for the magnetic field at the center of a coil that has N turns:
B = N mu0 I / 2 R = (500 turns) (4 pi X 10^-7 T m/A) (4.0 A) / 2 (10^-2 m)
= 0.12566 T = 0.126 T
(About five times greater than the publisher's answer)
Method 2
Using the solenoid formula, which textbooks describe as becoming more accurate as the windings become more closely packed, and as the length of the solenoid becomes much larger than its diameter, and if those assumptions are met then the formula expresses the nearly uniform field everywhere inside the solenoid, not only at the center:
turns per unit length = n = N/L = 500 turns / 0.10 m = 5000 turns per meter
B = mu0 n I = (4 pi X 10^-7 T m/A) (5000 turns per meter) (4.0 A)
= 0.025133 T = 0.025 T
(Agrees with the publisher's answer.)
Does the "coil" formula cease to become applicable just because the additional assumptions for using the "solenoid" formula have been provided? If so, why?
Thank you.
From Serway and Vuille, College Physics: "A solenoid with 500 turns, 0.10 m long, carrying a current of 4.0 A and with a radius of 10^-2 m will have what strength magnetic field at its center?" -- Answer provided by the publisher: 250 X 10^-4 T
My comparison of the two formulas ...
Method 1
Using the formula for the magnetic field at the center of a coil that has N turns:
B = N mu0 I / 2 R = (500 turns) (4 pi X 10^-7 T m/A) (4.0 A) / 2 (10^-2 m)
= 0.12566 T = 0.126 T
(About five times greater than the publisher's answer)
Method 2
Using the solenoid formula, which textbooks describe as becoming more accurate as the windings become more closely packed, and as the length of the solenoid becomes much larger than its diameter, and if those assumptions are met then the formula expresses the nearly uniform field everywhere inside the solenoid, not only at the center:
turns per unit length = n = N/L = 500 turns / 0.10 m = 5000 turns per meter
B = mu0 n I = (4 pi X 10^-7 T m/A) (5000 turns per meter) (4.0 A)
= 0.025133 T = 0.025 T
(Agrees with the publisher's answer.)
Does the "coil" formula cease to become applicable just because the additional assumptions for using the "solenoid" formula have been provided? If so, why?
Thank you.